A hydraulic turbine-generator unit placed at the bottom of a 75-m-high dam accepts water at a rate of 1020 L/s and produces 630 kWof electricity. Determine A) the overall efficiency of the turbine-generator unit and B) the turbine efficiency if the generator efficiency is 96 percent, and C) the power losses due to inefficiencies in the turbine and generator.

A. Overall efficiency = 83.95%

B. 87.45%

C. 120.465kw

Explanation:

H=75m

Flow rate = 120L/s=1.02m^2/s

Po=Output power = 630kw

As power input Pi = Total power available

=gQH/1000

Where

g=9.81

Q=120L/s=1.02m^2/s

H=75m

= (9.81m/s^2 * 1.02m^2/s*75m) / 1000

750.465kw

A. Overall efficiency of generator turbine unit (Po/Pi)

= Output power / power input

= 630/750.465 = 0.8395

Overall efficiency of the unit = 0.8395 * 100

=83.95%

B. Turbine efficiency if generator efficiency is 96%

no * nt = 0.96

nT = 87.45%

C. Power losses = Power input - Power output

Pi - Po

=750.465 - 630

=120.465kw

Power losses due to inefficiencies in turbine of the generator is 120.465kw